\(\left(x+y-2xy\right)\left(x+y+2xy\right)\)

\(=\left(x+y\right)^2-4x^2y^2\)

\(=x^2+2xy+y^2-4x^2y^2\)

\(\left(x+y-2xy\right)\left(x+y+2y\right)\)

\(=\left[\left(x+y\right)-2xy\right]\left[\left(x+y\right)+2xy\right]\)

\(=\left(x+y\right)^2-\left(2xy\right)^2\)

\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)

phân tích đa thức có dạng m² + n ( n thuộc z)

bàn làm giúp mình đk ko ạ!

\(a,-x^2y-2xy+2x^2y+5xy+2\\ =x^2y+3xy+2\\ b,-2xy+\dfrac{3}{2}xy^2+\dfrac{1}{2}xy^2+xy\\ =-xy+2xy^2\)

chọn D vì  \(\dfrac{7x-4}{2xy}+\dfrac{4-7x}{2xy}=0\)

a, \(\left(\frac{1}{2}xy-\frac{3}{4}\right)\left(\frac{1}{2}xy+\frac{3}{4}\right)=\frac{1}{4}x^2y^2+\frac{3}{8}xy-\frac{3}{8}xy-\frac{9}{16}=\frac{1}{4}x^2y^2-\frac{9}{16}\)

b, \(\left(2x+3\right)\left(4x^2-6x+9\right)=8x^3-12x^2+18x+12x^2-18x+27=8x^2+27\)

c, \(\left(xy-2\right)\left(x^2y^2+2xy+4\right)=x^3y^3+2x^2y^2+4xy-2x^2y^2-4xy-8=x^3y^3-8\)

Mk ko chép đề bài ra nhé

a, = 1/2xy.( -3/4 + 3/4 )

   =1/2xy.

b, Áp dụng HĐT số 2, có:  (Chỗ này ko cần chép cx đc)

=(2x+3).(2x+3)^2  (^ là mũ)

=(2x+3)^3

c, Áp dụng HĐT số 2, có:

=(xy-2).(xy + 2)^2

a) \(2x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow x^2+x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+10x+25\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x\\\left(x+5\right)^2\ge0\forall x\end{cases}}\)

\(\Rightarrow\left(x+y\right)^2+\left(x+5\right)^2\ge0\forall x\)

Vậy đẳng thức xảy ra\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=5\end{cases}}\)

b)\(x^2+3y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+y^2+2y^2+2xy-2y+\frac{1}{2}+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(2y^2-2y+\frac{1}{2}\right)+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)

Vì \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2\ge0\)

nên \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

Mà\(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)

nên pt vô nghiệm

`x^2 -4x+4-y^2`

`=(x^2 -4x+4)-y^2`

`=(x-2)^2 -y^2`

`=(x-2-y)(x-2+y)`

`x^2+2xy+y^2-x-y`

`=(x^2+2xy+y^2) -(x+y)`

`=(x+y)^2 -(x+y)`

`=(x+y)(x+y-1)`

`x^2-2xy+y^2-9`

`=(x^2-2xy+y^2)-3^2`

`=(x-y)^2-3^3`

`=(x-y-3)(x-y+3)`

Tách ra đi cậu.

https://hoc24.vn/cau-hoi/927x118.8505894378996

giúp mik với ạ

2x³y - 2xy³ - 4xy² -2xy

=2xy.(x²-y²-2y-1)

=2xy.[x²-(y²+2x+1)]

=2xy.[x²-(y+1)²]

=2xy.[x+(y+1)][x-(y+1)]

=2xy.(x+y+1)(x-y-1)